Section 3.4: (Discrete) Random Variables

Section 3.4: (Discrete) Random Variables

1 Two Examples

De…nition 1 A random variable is a function that assigns a numerical value to each outcome in a sample space.

There are two types of random variables, discrete and continuous, based on the data types assigned to values. The above two examples illustrate the di¤er- ent ways a random variable must be de…ned. Accordingly, di¤erent techniques will be needed to work with the two di¤erent types of random variables.

Example 2 Consider the experiment of rolling a pair of dice and summing the faces. The random variable X assigns to each roll its sum. The following table indicates the probabilities for each value in X.

Sum 2 3 4 5 6 7 8 9 10 11 12

Remark 3 Since the data is discrete, we call this function a discrete random variable.

Example 4 Let Y be the random variable that assigns to each human being, their height in inches. Given the uncountably in…nite number of possible heights, it is not possible to de…ne Y based on a table.

Remark 5 Since the data is continuous, we call this function a continuous random variable.

Problem 6 Consider the experiment of rolling a three sided-die and a four- sided die and summing the faces. Construct a discrete random variable X.

De…nition 7 A random variable with only two outcomes (0-1, true-false, right- wrong, on-o¤ , etc.) is called a Bernoulli random variable.

Example 8 A multiple choice test with only one correct answer per question is a Bernoulli random variable.

Problem 9 Formally describe a public safety o¢ cer’s parking lot duty as a random variable.

De…nition 10 A probability density function (pdf ) or probability mass function (pmf ) for a discrete random variable X is a function whose domain is all possible values of X and assigns to each x 2 X the probability that x occurs.

Note that the sum of all probabilities in a distribution must be 1.

Problem 11 Is the following function a probability distribution? Explain.

X 1 2 3

P (X) 0.5 0.6 -0.1

Example 12 Consider the following probability distribution.

X 1 2 3 4 5 6

P (X) 0.3 0.2 0.1 0.1 0.2 1. P (X = 3) =

2. P (X = 4) = 3. P (X = 5) = 4. P (X 3) = 5. P (X < 3) = 6. P (2 X 4) = 7. P (X > 5) =

Example 13 Consider the experiment of rolling a pair of dice and summing the faces. The random variable X assigns to each roll its sum. The following table indicates the probabilities for each value in X. Together they form the probability density function.

Sum 2 3 4 5 6 7 8 9 10 11 12

Prob ₃₆¹ ₃₆² ₃₆³ ₃₆⁴ ₃₆⁵ ₃₆⁶ ₃₆⁵ ₃₆⁴ ₃₆³ ₃₆² ₃₆¹

Problem 14 Consider the experiment of rolling a three sided-die and a 4-sided die and summing the faces. Construct the pdf for X.

Problem 15 A $2 lottery ticket o¤ ers four chances to win di¤ erent amounts of money as indicated by the following probability distribution model.

Prize $0 $0.5 $1 $5 $50 Prob ₁₀₀⁷¹ ₁₀₀¹⁵ ₁₀₀¹⁰ ₁₀₀³ ₁₀₀¹ 1. What is the probability that a ticket wins nothing?

2. What is the probability that a ticket wins money but is a losing proposition overall?

3. Two people buy one ticket each. What is the probability that they both win nothing?

4. Two people buy one ticket each. What is the probability that they both win money?

5. Ten people buy one ticket each. What is the probability that at least one person wins money?

6. Five hundred people buy one ticket each. What is the probability that at least one person wins the $50 prize?

Problem 16 You draw a card from a deck. If you get a club you get $5. If you get an Ace you get $10. For all other cards you receive nothing. Let X be the random variable of money won. Create a probability density function for this game.

Problem 17 You draw a card from a deck. If you get a club you get nothing.

If you get red card you get $10. If you get a spade you get $15 and get to select another card (without replacement). If the second is another spade you receive an additional $20. You receive nothing for any non-spade card. Create a probability density function for this game.

A random variable with a …nite number of outcomes is necessarily discrete.

However, it is possible for a discrete random variable to have an in…nite number of outcomes.

Example 18 John ‡ips a coin until he observes a tail. Construct the pdf for the number of ‡ips of the coin for this discrete random variable.

2 Cumulative Distribution Functions

The cumulative distribution function of a discrete pdf is the sum of all proba- bilities of outcomes less than or equal to some …xed value x.

De…nition 19 The cumulative distribution function F(x) of a pdf p(x) is P

y x

p(y).

Example 20 Consider the experiment of rolling a pair of dice and summing the faces. The random variable X assigns to each roll its sum. The following table indicates the probabilities for each value in X. Together they form the probability density function.

Sum 2 3 4 5 6 7 8 9 10 11 12

p(x) ₃₆¹ ₃₆² ₃₆³ ₃₆⁴ ₃₆⁵ ₃₆⁶ ₃₆⁵ ₃₆⁴ ₃₆³ ₃₆² ₃₆¹ Now construct the cumulative distribution function of this pdf.

Sum 2 3 4 5 6 7 8 9 10 11 12

F (x) ₃₆¹ ₃₆³ ₃₆^{6 10}₃₆ ¹⁵₃₆ ²¹₃₆ ²⁶₃₆ ³⁰₃₆ ³³₃₆ ³⁵₃₆ ³⁶₃₆

Problem 21 Consider the experiment of rolling a three sided-die and a 4-sided die and summing the faces. Construct p(x) and F (x) for X.

Example 22 Consider rolling a fair die until a 5 appears. Construct the pdf for the number of rolls.

The probability of rolling a 5 (or any speci…c value) on a fair die is p = ¹₆. Since there is never a guarantee of ever getting a 5, the set of possible values for x is the set of positive integers. Since Z⁺ is in…nite, one cannot write down all possible values and probabilities. Instead we need to construct a function. It can help to model a few speci…c values of x before writing a general form. For example, the …rst 5 occurs on attempt number four with probability ⁵₆ ^{3 1}₆ =

125

1296. One has to roll a number other than 5, three times and then roll a 5.

Generalizing this our pdf, p(x) = ⁵₆ ^{x 1 1}₆.

Expected Value (Mean) and Standard Devia- tion for a Discrete Random Variable

Recall the experiment of rolling a pair of dice and summing the faces. The random variable X assigns to each roll its sum. The following table indicates the probabilities for each value in X . Together they form the probability density

The pdf provides an easy table or function from which to acquire probabilities of individual roles. What value on average do we expect after rolling the pair of dice? What value on average do we expect after rolling the pair of dice 100 times? How much variation do we expect from that average roll? These sound like questions of a mean and standard deviation. Rather than being a mean and standard deviation of a data set, the pdf introduces the notion of uncertainty and likelihood.

De…nition 23 The expected value (or mean), E(X)(or ); of a discrete probability distribution is given by

E(X) = X

x2X

x p(x).

De…nition 24 The variance, V (X) ; of a discrete probability distribution is given by

2= X

x2X

(x E(X))² p(x).

Example 25 Back to the experiment of rolling a pair of dice and summing the faces. The random variable X assigns to each roll its sum. The following table is our previously constructed probability model.

Sum 2 3 4 5 6 7 8 9 10 11 12

Prob ₃₆¹ ₃₆² ₃₆³ ₃₆⁴ ₃₆⁵ ₃₆⁶ ₃₆⁵ ₃₆⁴ ₃₆³ ₃₆² ₃₆¹

What is the expected value of the sum of a pair of dice?

E(X) = X

x2X

x p(x)

= 2 1

36+ 3 2

36+ 4 3

36+ 5 4

36+ 6 5

36+ 7 6 36 +8 5

36+ 9 4

36+ 10 3

36+ 11 2

36+ 12 1 36

= 2 + 6 + 12 + 20 + 30 + 42 + 40 + 36 + 30 + 22 + 12 36

= 7.

What is the standard deviation of the sum of a pair of dice?

Sum 2 3 4 5 6 7 8 9 10 11 12

(x E(X))² 25 16 9 4 1 0 1 4 9 16 25

Prob ₃₆¹ ₃₆² ₃₆³ ₃₆⁴ ₃₆⁵ ₃₆⁶ ₃₆⁵ ₃₆⁴ ₃₆³ ₃₆² ₃₆¹

V (X) = ²= X

x2X

(x E(X))² p(x)

= 25 1

36+ 16 2

36+ 9 3

36+ 4 4

36+ 1 5

36+ 0 6 36 +1 5

36+ 4 4

36+ 9 3

36+ 16 2

36+ 25 1 36

= 35 6 . Thus, =q

35

6 = 2: 415 2.

Example 26 What is the probability that the sum of two dice falls within one standard deviation of the mean?

One standard deviation from the mean is (7 2:4152) = (4: 584 8; 9: 415 2). So we could see a sum of 5, 6, 7, 8, or 9. This occurs with probability

p = 4 36+ 5

36+ 6 36+ 5

36+ 4 36 =2

3 = 0:666 67:

Exercise 27 A $2 lottery ticket o¤ ers four chances to win di¤ erent amounts of money as indicated by the following probability distribution model. Determine the expected value of buying a single ticket. Do you wish to play this game?

Prize $0 $0.5 $1 $5 $50

Prob ₁₀₀⁷¹ ₁₀₀¹⁵ ₁₀₀¹⁰ ₁₀₀³ ₁₀₀¹

Value 2 1:5 1 3 48

So,

E(X) = X

x2X

x p(x)

= 2 71

100 1:5 15

100 1 10

100 + 3 3

100 + 48 1 100

= 1: 175.

I personally would not want to play this game with a negative expected value and such a small prize.

Exercise 28 What is the expected value of buying 50 such lottery tickets?

1: 175 50 = 58: 75.

Problem 29 You pay $1 to play a game. The game consists of rolling a pair of dice. If you observe a sum of 7 or 11 you receive $4. If not, you receive nothing. Compute the expected value for this game?

Example 30 Consider ‡ipping an unbalanced coin that lands H 60% of the time. What is the expected value and standard deviation for this Bernoulli ran- dom variable?

Coin H T

Prob :6 :4 is the pdf. It’s expected value is that the coin lands H, 60% of the time. One can view this representing a success with a 1 and a failure as a 0 for the X values.

X 1 0

p(x) :6 :4

E(X) = X

x2X

x p(x)

= 1 :6 + 0 :4

= :6

2 = X

x2X

(x E(X))² p(x)

= (1 :6)² :6 + (0 :6)² :4

= :24

Thus, = p

:24

= 0:489 90

While not immediately obvious, it should be noted that ²= :6 :4.

At times we will want to use a given expectation as an input variable into some other function.

Example 31 Our electronics store buys 5 game consoles at $250 each. The pdf for the number of consoles sold in the …rst week is given below.

# sold 0 1 2 3 4 5

p(x) .1 .1 .2 .3 .2 .1

We expect to sell 0 :1 + 1 :1 + 2 :2 + 3 :3 + 4 :2 + 5 :1 = 2: 7 consoles this week.

This week the console is hot and can be sold for $350 each. After this week, we must sell the remaining consoles at $200 each. What is our expected pro…t? If x represents the number of consoles sold then our pro…t function

E(f (x)) = X5 i=0

f (i) p(i)

= f (0) p(0) + f (1) p(1) + f (2) p(2) +f (3) p(3) + f (4) p(4) + f (5) p(5)

= 250 :1 100 :1 + 50 :2 + 200 :3 + 350 :2 + 500 :1

= 155:

Remark 32 Note with wonder and amazement that 150 2:7 250 = 155:0:

3 Exercises

Navidi Section 2.4: 1-10